Question

$y_1=88sin(\omega t-kx)$ and $y_2=6sin(\omega t+kx)$ are two waves travelling in a string of area of cross-section $s$ and density $\rho$. These two waves are superimposed to produce a standing wave. Find the total amount of energy crossing through a node per second.

• A. $\frac{2\rho {\omega }^{3}s}{k}$
• B. $\frac{3\rho {\omega }^{3}s}{k}$
• C. $\frac{5\rho {\omega }^{3}s}{k}$
• D. $\frac{6\rho {\omega }^{3}s}{k}$

Answer 1

The correct option is A $\frac{2\rho {\omega }^{3}s}{k}$

Here, we will take ,

$y=y_1+y_2$

= $8sin(\omega t-kx)+6sin(\omega t+kx)$..........(1)

Now, add and subtract $6sin(\omega t-kx)$ in (1).

= $8sin(\omega t-kx)+6sin(\omega t+kx)+6sin(\omega t-kx)-6sin(\omega t-kx)$

=$2sin(\omega t-kx)+12sin\omega tcoskx$............(2)

We obtained (2) by solving using trigonometric relations.

energy crossing through node per second = power

$P=\frac{1}{2}\rho {\omega }^2(2{)}^{2}Sv$..................(3)

Now, put $v=\frac{\omega }{k}$ in eqn. (3)

We get , $P=\frac{2\rho {\omega }^{3}s}{k}$